# Winding number of a simple closed curve in a plane.

I know that a simple closed curve $\gamma$ on a plane divides the plane into two connected components, by the Jordan curve theorem for example. If I use the definition $$\text{Winding number }(a,\gamma)=\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z-a},$$ then this is a continuous function in $a$ on $\mathbb{C}\backslash\gamma$. Using the fact that the integral only gives integer values, I know that the integral should be 0 on the unbounded component of $\mathbb{C}\backslash\gamma$. The integral should also be constant on the other connected component, but how does one prove that the value has to be 1 or -1?

• (Of course $\gamma$ has to be rectifiable for your integral to make sense.) Suppose for convenience of notation that $0$ is in the bounded component. Use Green's theorem to show that, because $\gamma$ is homotopic to the standard unit circle (or that with the opposite orientation, depending on how $\gamma$ is oriented), this integral is the same as the integral over the unit circle. Then integrate just the one. – user98602 Mar 15 '16 at 5:43
• @Mike Miller: Can I ask how to know $\gamma$ is homotopic to the standard unit circle? If I just use $h(s,t)=s\gamma(t)+(1-s)C(t)$, where $C(t)$ is the unit circle, it seemed that this might sweep through the origin. – DY88 Mar 15 '16 at 11:15
• @MIke Miller: Ah. Maybe I don't need to construct homotopy but just draw small circle about 0 that doesn't intersect the simple closed curve $\gamma$ and from some two different points $a,b$ on the unit circle, draw two line segments that connects to $\hat{a},\hat{b}$ on $\gamma$ respectively but such that the two line segments only intersect $\gamma$ at $\hat{a},\hat{b}$. Then I would get two simple closed curve and neither doesn't contain 0 in the inner region and the integral should be 0 on the two curves, with appropriate orientation given on the circle and the added segments. – DY88 Mar 15 '16 at 12:17